Advanced Calculus Single Variable

Chapter 6Continuous Functions

The concept of function is far too general to be useful in calculus. There are various ways to
restrict the concept in order to study something interesting and the types of restrictions
considered depend very much on what you find interesting. In Calculus, the most fundamental
restriction made is to assume the functions are continuous. Continuous functions are those in
which a sufficiently small change in x results in a small change in f

(x)

. They rule out things
which could never happen physically. For example, it is not possible for a car to jump from
one point to another instantly. Making this restriction precise turns out to be surprisingly
difficult although many of the most important theorems about continuous functions seem
intuitively clear.

Before giving the careful mathematical definitions, here are examples of graphs of
functions which are not continuous at the point x0.

PICT

You see, there is a hole in the picture of the graph of this function and instead of filling in
the hole with the appropriate value, f

(x0)

is too large. This is called a removable
discontinuity because the problem can be fixed by redefining the function at the point x0.
Here is another example.

PICT

You see from this picture that there is no way to get rid of the jump in the graph of
this function by simply redefining the value of the function at x0. That is why it
is called a nonremovable discontinuity or jump discontinuity. Now that pictures
have been given of what it is desired to eliminate, it is time to give the precise
definition.

The definition which follows, due to
Cauchy1 and
Weierstrass2 is the precise way to exclude the sort of behavior described above and all statements about
continuous functions must ultimately rest on this definition from now on.

Definition 6.0.1A function f : D

(f)

⊆ F → F is continuous at x ∈ D

(f)

if foreach ε > 0 there exists δ > 0 such that whenever y ∈ D

(f)

and

|y− x| < δ

it follows that

|f (x)− f (y)| < ε.

A function f is continuous if it is continuous at every point of D

(f)

.

In sloppy English this definition says roughly the following: A function f is continuous at
x when it is possible to make f

(y)

as close as desired to f

(x)

provided y is taken close
enough to x. In fact this statement in words is pretty much the way Cauchy described it. The
completely rigorous definition above is due to Weierstrass. This definition does indeed rule out
the sorts of graphs drawn above. Consider the second nonremovable discontinuity. The
removable discontinuity case is similar.

PICT

For the ε shown you can see from the picture that no matter how small you take δ, there
will be points, x, between x0−δ and x0 where f

(x)

< 2 −ε. In particular, for these values of
x,

|f (x)− f (x0)|

> ε. Therefore, the definition of continuity given above excludes the
situation in which there is a jump in the function. Similar reasoning shows it excludes the
removable discontinuity case as well. There are many ways a function can fail to be
continuous and it is impossible to list them all by drawing pictures. This is why it is so
important to use the definition. The other thing to notice is that the concept of
continuity as described in the definition is a point property. That is to say it is
a property which a function may or may not have at a single point. Here is an
example.

Example 6.0.2Let

{
x if x is rational
f (x) = 0 if x is irrational .

This function is continuous at x = 0 and nowhere else.

To verify the assertion about the above function, first show it is not continuous at x if
x≠0. Take such an x and let ε =

|x|

∕2. Now let δ > 0 be completely arbitrary. In the interval,

(x − δ,x + δ)

there are rational numbers, y1 such that

|y1|

>

|x|

and irrational numbers, y2.
Thus

|f (y1)− f (y2)|

=

|y1|

>

|x|

. If f were continuous at x, there would exist δ > 0 such
that for every point, y ∈

(x − δ,x + δ)

,

|f (y)− f (x)|

< ε. But then, letting y1 and y2 be as
just described,

|x|

<

|y1|

=

|f (y1)− f (y2)|

≤

|f (y1)− f (x )|

+

|f (x)− f (y2)|

< 2ε =

|x|

,

which is a contradiction. Since a contradiction is obtained by assuming that f
is continuous at x, it must be concluded that f is not continuous there. To see
f is continuous at 0, let ε > 0 be given and let δ = ε. Then if

|y− 0|

< δ = ε,
Then

|f (y) − f (0)|

= 0 if y is irrational

|f (y) − f (0)|

=

|y|

< ε if y is rational.

either way, whenever

|y− 0|

< δ, it follows

|f (y) − f (0)|

< ε and so f is continuous at
x = 0. How did I know to let δ = ε? That is a very good question. The choice of δ for a
particular ε is usually arrived at by using intuition, the actual εδ argument reduces to a
verification that the intuition was correct. Here is another example.

Example 6.0.3Show the function f

(x)

= −5x + 10 is continuous at x = −3.

To do this, note first that f

(− 3)

= 25 and it is desired to verify the conditions for
continuity. Consider the following.

|− 5x + 10 − (25)| = 5|x − (− 3)|.

This allows one to find a suitable δ. If ε > 0 is given, let 0 < δ ≤

15

ε. Then if 0 <

|x − (− 3)|

< δ,
it follows from this inequality that

|− 5x+ 10− (25)| = 5|x− (− 3)| < 51ε = ε.
5

Sometimes the determination of δ in the verification of continuity can be a little more
involved. Here is another example.